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It is given that for a finite population of 2500 women, 82% of them give birth.

The confidence level is given to be equal to 99%, and the margin of error is equal to 0.02 (two percentage points).

The level of significance is equal to 0.01.

The value of $z_{\frac{伪}{2}}$ from the standard normal distribution table is equal to 2.5758.

Use the given formula to compute the sample size:

Here,

$\hat{p}$ is the sample proportion of women who have given birth and is equal to 0.82.

$\hat{q}$ is the sample proportion of women who have not given birth.

N is the population size and is equal to 2500.

E is the margin of error is given to be equal to 0.02.

$z_{\frac{伪}{2}}$is equal to 2.5758.

Substitute all the values in the given formula as follows:

The formula for sample size,

$$\begin{gathered} n=\frac{N脳\hat{p}\hat{q}脳{\left(z_{\frac{伪}{2}}\right)}^2}{\hat{p}\hat{q}脳{\left(z_{\frac{伪}{2}}\right)}^2+\left(N-1\right)E^2} \\ =\frac{2500脳\left(0.82脳\left(1-0.82\right)\right)脳{\left(2.5758\right)}^2}{\left(0.82脳\left(1-0.82\right)\right)脳{\left(2.5758\right)}^2+\left(2500-1\right)脳{\left(0.02\right)}^2} \\ 鈮�1238 \end{gathered}$$

Therefore, the sample size required is equal to 1238.